3.337 \(\int \frac{x^4}{(3+2 x^2) \sqrt{1+2 x^2+2 x^4}} \, dx\)

Optimal. Leaf size=418 \[ \frac{\left (1-3 \sqrt{2}\right ) \left (\sqrt{2} x^2+1\right ) \sqrt{\frac{2 x^4+2 x^2+1}{\left (\sqrt{2} x^2+1\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}\left (\sqrt [4]{2} x\right ),\frac{1}{4} \left (2-\sqrt{2}\right )\right )}{2\ 2^{3/4} \left (2-3 \sqrt{2}\right ) \sqrt{2 x^4+2 x^2+1}}+\frac{\sqrt{2 x^4+2 x^2+1} x}{2 \sqrt{2} \left (\sqrt{2} x^2+1\right )}-\frac{3 \sqrt{\frac{3}{10}} \left (3-\sqrt{2}\right ) \tan ^{-1}\left (\frac{\sqrt{\frac{5}{3}} x}{\sqrt{2 x^4+2 x^2+1}}\right )}{4 \left (2-3 \sqrt{2}\right )}-\frac{\left (\sqrt{2} x^2+1\right ) \sqrt{\frac{2 x^4+2 x^2+1}{\left (\sqrt{2} x^2+1\right )^2}} E\left (2 \tan ^{-1}\left (\sqrt [4]{2} x\right )|\frac{1}{4} \left (2-\sqrt{2}\right )\right )}{2\ 2^{3/4} \sqrt{2 x^4+2 x^2+1}}+\frac{3 \left (3+\sqrt{2}\right ) \left (\sqrt{2} x^2+1\right ) \sqrt{\frac{2 x^4+2 x^2+1}{\left (\sqrt{2} x^2+1\right )^2}} \Pi \left (\frac{1}{24} \left (12-11 \sqrt{2}\right );2 \tan ^{-1}\left (\sqrt [4]{2} x\right )|\frac{1}{4} \left (2-\sqrt{2}\right )\right )}{8\ 2^{3/4} \left (2-3 \sqrt{2}\right ) \sqrt{2 x^4+2 x^2+1}} \]

[Out]

(x*Sqrt[1 + 2*x^2 + 2*x^4])/(2*Sqrt[2]*(1 + Sqrt[2]*x^2)) - (3*Sqrt[3/10]*(3 - Sqrt[2])*ArcTan[(Sqrt[5/3]*x)/S
qrt[1 + 2*x^2 + 2*x^4]])/(4*(2 - 3*Sqrt[2])) - ((1 + Sqrt[2]*x^2)*Sqrt[(1 + 2*x^2 + 2*x^4)/(1 + Sqrt[2]*x^2)^2
]*EllipticE[2*ArcTan[2^(1/4)*x], (2 - Sqrt[2])/4])/(2*2^(3/4)*Sqrt[1 + 2*x^2 + 2*x^4]) + ((1 - 3*Sqrt[2])*(1 +
Sqrt[2]*x^2)*Sqrt[(1 + 2*x^2 + 2*x^4)/(1 + Sqrt[2]*x^2)^2]*EllipticF[2*ArcTan[2^(1/4)*x], (2 - Sqrt[2])/4])/(
2*2^(3/4)*(2 - 3*Sqrt[2])*Sqrt[1 + 2*x^2 + 2*x^4]) + (3*(3 + Sqrt[2])*(1 + Sqrt[2]*x^2)*Sqrt[(1 + 2*x^2 + 2*x^
4)/(1 + Sqrt[2]*x^2)^2]*EllipticPi[(12 - 11*Sqrt[2])/24, 2*ArcTan[2^(1/4)*x], (2 - Sqrt[2])/4])/(8*2^(3/4)*(2
- 3*Sqrt[2])*Sqrt[1 + 2*x^2 + 2*x^4])

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Rubi [A]  time = 0.184204, antiderivative size = 418, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.138, Rules used = {1325, 1103, 1195, 1706} \[ \frac{\sqrt{2 x^4+2 x^2+1} x}{2 \sqrt{2} \left (\sqrt{2} x^2+1\right )}-\frac{3 \sqrt{\frac{3}{10}} \left (3-\sqrt{2}\right ) \tan ^{-1}\left (\frac{\sqrt{\frac{5}{3}} x}{\sqrt{2 x^4+2 x^2+1}}\right )}{4 \left (2-3 \sqrt{2}\right )}+\frac{\left (1-3 \sqrt{2}\right ) \left (\sqrt{2} x^2+1\right ) \sqrt{\frac{2 x^4+2 x^2+1}{\left (\sqrt{2} x^2+1\right )^2}} F\left (2 \tan ^{-1}\left (\sqrt [4]{2} x\right )|\frac{1}{4} \left (2-\sqrt{2}\right )\right )}{2\ 2^{3/4} \left (2-3 \sqrt{2}\right ) \sqrt{2 x^4+2 x^2+1}}-\frac{\left (\sqrt{2} x^2+1\right ) \sqrt{\frac{2 x^4+2 x^2+1}{\left (\sqrt{2} x^2+1\right )^2}} E\left (2 \tan ^{-1}\left (\sqrt [4]{2} x\right )|\frac{1}{4} \left (2-\sqrt{2}\right )\right )}{2\ 2^{3/4} \sqrt{2 x^4+2 x^2+1}}+\frac{3 \left (3+\sqrt{2}\right ) \left (\sqrt{2} x^2+1\right ) \sqrt{\frac{2 x^4+2 x^2+1}{\left (\sqrt{2} x^2+1\right )^2}} \Pi \left (\frac{1}{24} \left (12-11 \sqrt{2}\right );2 \tan ^{-1}\left (\sqrt [4]{2} x\right )|\frac{1}{4} \left (2-\sqrt{2}\right )\right )}{8\ 2^{3/4} \left (2-3 \sqrt{2}\right ) \sqrt{2 x^4+2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

Int[x^4/((3 + 2*x^2)*Sqrt[1 + 2*x^2 + 2*x^4]),x]

[Out]

(x*Sqrt[1 + 2*x^2 + 2*x^4])/(2*Sqrt[2]*(1 + Sqrt[2]*x^2)) - (3*Sqrt[3/10]*(3 - Sqrt[2])*ArcTan[(Sqrt[5/3]*x)/S
qrt[1 + 2*x^2 + 2*x^4]])/(4*(2 - 3*Sqrt[2])) - ((1 + Sqrt[2]*x^2)*Sqrt[(1 + 2*x^2 + 2*x^4)/(1 + Sqrt[2]*x^2)^2
]*EllipticE[2*ArcTan[2^(1/4)*x], (2 - Sqrt[2])/4])/(2*2^(3/4)*Sqrt[1 + 2*x^2 + 2*x^4]) + ((1 - 3*Sqrt[2])*(1 +
Sqrt[2]*x^2)*Sqrt[(1 + 2*x^2 + 2*x^4)/(1 + Sqrt[2]*x^2)^2]*EllipticF[2*ArcTan[2^(1/4)*x], (2 - Sqrt[2])/4])/(
2*2^(3/4)*(2 - 3*Sqrt[2])*Sqrt[1 + 2*x^2 + 2*x^4]) + (3*(3 + Sqrt[2])*(1 + Sqrt[2]*x^2)*Sqrt[(1 + 2*x^2 + 2*x^
4)/(1 + Sqrt[2]*x^2)^2]*EllipticPi[(12 - 11*Sqrt[2])/24, 2*ArcTan[2^(1/4)*x], (2 - Sqrt[2])/4])/(8*2^(3/4)*(2
- 3*Sqrt[2])*Sqrt[1 + 2*x^2 + 2*x^4])

Rule 1325

Int[(x_)^4/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4]), x_Symbol] :> With[{q = Rt[c/a, 2]
}, -Dist[(2*c*d - a*e*q)/(c*e*(e - d*q)), Int[1/Sqrt[a + b*x^2 + c*x^4], x], x] + (-Dist[1/(e*q), Int[(1 - q*x
^2)/Sqrt[a + b*x^2 + c*x^4], x], x] + Dist[d^2/(e*(e - d*q)), Int[(1 + q*x^2)/((d + e*x^2)*Sqrt[a + b*x^2 + c*
x^4]), x], x])] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && PosQ[c/a] && NeQ[c*d^2 - a*e^2, 0]

Rule 1103

Int[1/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, Simp[((1 + q^2*x^2)*Sqrt[(
a + b*x^2 + c*x^4)/(a*(1 + q^2*x^2)^2)]*EllipticF[2*ArcTan[q*x], 1/2 - (b*q^2)/(4*c)])/(2*q*Sqrt[a + b*x^2 + c
*x^4]), x]] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && PosQ[c/a]

Rule 1195

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, -Simp[
(d*x*Sqrt[a + b*x^2 + c*x^4])/(a*(1 + q^2*x^2)), x] + Simp[(d*(1 + q^2*x^2)*Sqrt[(a + b*x^2 + c*x^4)/(a*(1 + q
^2*x^2)^2)]*EllipticE[2*ArcTan[q*x], 1/2 - (b*q^2)/(4*c)])/(q*Sqrt[a + b*x^2 + c*x^4]), x] /; EqQ[e + d*q^2, 0
]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && PosQ[c/a]

Rule 1706

Int[((A_) + (B_.)*(x_)^2)/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4]), x_Symbol] :> With[
{q = Rt[B/A, 2]}, -Simp[((B*d - A*e)*ArcTan[(Rt[-b + (c*d)/e + (a*e)/d, 2]*x)/Sqrt[a + b*x^2 + c*x^4]])/(2*d*e
*Rt[-b + (c*d)/e + (a*e)/d, 2]), x] + Simp[((B*d + A*e)*(A + B*x^2)*Sqrt[(A^2*(a + b*x^2 + c*x^4))/(a*(A + B*x
^2)^2)]*EllipticPi[Cancel[-((B*d - A*e)^2/(4*d*e*A*B))], 2*ArcTan[q*x], 1/2 - (b*A)/(4*a*B)])/(4*d*e*A*q*Sqrt[
a + b*x^2 + c*x^4]), x]] /; FreeQ[{a, b, c, d, e, A, B}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^
2, 0] && NeQ[c*d^2 - a*e^2, 0] && PosQ[c/a] && EqQ[c*A^2 - a*B^2, 0]

Rubi steps

\begin{align*} \int \frac{x^4}{\left (3+2 x^2\right ) \sqrt{1+2 x^2+2 x^4}} \, dx &=-\frac{\int \frac{1-\sqrt{2} x^2}{\sqrt{1+2 x^2+2 x^4}} \, dx}{2 \sqrt{2}}+\frac{9 \int \frac{1+\sqrt{2} x^2}{\left (3+2 x^2\right ) \sqrt{1+2 x^2+2 x^4}} \, dx}{2 \left (2-3 \sqrt{2}\right )}-\frac{\left (12-2 \sqrt{2}\right ) \int \frac{1}{\sqrt{1+2 x^2+2 x^4}} \, dx}{4 \left (2-3 \sqrt{2}\right )}\\ &=\frac{x \sqrt{1+2 x^2+2 x^4}}{2 \sqrt{2} \left (1+\sqrt{2} x^2\right )}-\frac{3 \sqrt{\frac{3}{10}} \left (3-\sqrt{2}\right ) \tan ^{-1}\left (\frac{\sqrt{\frac{5}{3}} x}{\sqrt{1+2 x^2+2 x^4}}\right )}{4 \left (2-3 \sqrt{2}\right )}-\frac{\left (1+\sqrt{2} x^2\right ) \sqrt{\frac{1+2 x^2+2 x^4}{\left (1+\sqrt{2} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\sqrt [4]{2} x\right )|\frac{1}{4} \left (2-\sqrt{2}\right )\right )}{2\ 2^{3/4} \sqrt{1+2 x^2+2 x^4}}+\frac{\left (1-3 \sqrt{2}\right ) \left (1+\sqrt{2} x^2\right ) \sqrt{\frac{1+2 x^2+2 x^4}{\left (1+\sqrt{2} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\sqrt [4]{2} x\right )|\frac{1}{4} \left (2-\sqrt{2}\right )\right )}{2\ 2^{3/4} \left (2-3 \sqrt{2}\right ) \sqrt{1+2 x^2+2 x^4}}+\frac{3 \left (3+\sqrt{2}\right ) \left (1+\sqrt{2} x^2\right ) \sqrt{\frac{1+2 x^2+2 x^4}{\left (1+\sqrt{2} x^2\right )^2}} \Pi \left (\frac{1}{24} \left (12-11 \sqrt{2}\right );2 \tan ^{-1}\left (\sqrt [4]{2} x\right )|\frac{1}{4} \left (2-\sqrt{2}\right )\right )}{8\ 2^{3/4} \left (2-3 \sqrt{2}\right ) \sqrt{1+2 x^2+2 x^4}}\\ \end{align*}

Mathematica [C]  time = 0.218542, size = 127, normalized size = 0.3 \[ -\frac{\sqrt{1+(1-i) x^2} \sqrt{1+(1+i) x^2} \left (-(1+4 i) \text{EllipticF}\left (i \sinh ^{-1}\left (\sqrt{1-i} x\right ),i\right )+(1+i) E\left (\left .i \sinh ^{-1}\left (\sqrt{1-i} x\right )\right |i\right )+3 i \Pi \left (\frac{1}{3}+\frac{i}{3};\left .i \sinh ^{-1}\left (\sqrt{1-i} x\right )\right |i\right )\right )}{4 \sqrt{1-i} \sqrt{2 x^4+2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^4/((3 + 2*x^2)*Sqrt[1 + 2*x^2 + 2*x^4]),x]

[Out]

-(Sqrt[1 + (1 - I)*x^2]*Sqrt[1 + (1 + I)*x^2]*((1 + I)*EllipticE[I*ArcSinh[Sqrt[1 - I]*x], I] - (1 + 4*I)*Elli
pticF[I*ArcSinh[Sqrt[1 - I]*x], I] + (3*I)*EllipticPi[1/3 + I/3, I*ArcSinh[Sqrt[1 - I]*x], I]))/(4*Sqrt[1 - I]
*Sqrt[1 + 2*x^2 + 2*x^4])

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Maple [C]  time = 0.02, size = 222, normalized size = 0.5 \begin{align*}{\frac{ \left ( -{\frac{1}{4}}+{\frac{i}{4}} \right ) \left ({\it EllipticF} \left ( x\sqrt{-1+i},{\frac{\sqrt{2}}{2}}+{\frac{i}{2}}\sqrt{2} \right ) -{\it EllipticE} \left ( x\sqrt{-1+i},{\frac{\sqrt{2}}{2}}+{\frac{i}{2}}\sqrt{2} \right ) \right ) }{\sqrt{-1+i}}\sqrt{1+ \left ( 1-i \right ){x}^{2}}\sqrt{1+ \left ( 1+i \right ){x}^{2}}{\frac{1}{\sqrt{2\,{x}^{4}+2\,{x}^{2}+1}}}}-{\frac{3\,{\it EllipticF} \left ( x\sqrt{-1+i},1/2\,\sqrt{2}+i/2\sqrt{2} \right ) }{4\,\sqrt{-1+i}}\sqrt{1+ \left ( 1-i \right ){x}^{2}}\sqrt{1+ \left ( 1+i \right ){x}^{2}}{\frac{1}{\sqrt{2\,{x}^{4}+2\,{x}^{2}+1}}}}+{\frac{3}{4\,\sqrt{-1+i}}\sqrt{-i{x}^{2}+{x}^{2}+1}\sqrt{i{x}^{2}+{x}^{2}+1}{\it EllipticPi} \left ( x\sqrt{-1+i},{\frac{1}{3}}+{\frac{i}{3}},{\frac{\sqrt{-1-i}}{\sqrt{-1+i}}} \right ){\frac{1}{\sqrt{2\,{x}^{4}+2\,{x}^{2}+1}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/(2*x^2+3)/(2*x^4+2*x^2+1)^(1/2),x)

[Out]

(-1/4+1/4*I)/(-1+I)^(1/2)*(1+(1-I)*x^2)^(1/2)*(1+(1+I)*x^2)^(1/2)/(2*x^4+2*x^2+1)^(1/2)*(EllipticF(x*(-1+I)^(1
/2),1/2*2^(1/2)+1/2*I*2^(1/2))-EllipticE(x*(-1+I)^(1/2),1/2*2^(1/2)+1/2*I*2^(1/2)))-3/4/(-1+I)^(1/2)*(1+(1-I)*
x^2)^(1/2)*(1+(1+I)*x^2)^(1/2)/(2*x^4+2*x^2+1)^(1/2)*EllipticF(x*(-1+I)^(1/2),1/2*2^(1/2)+1/2*I*2^(1/2))+3/4/(
-1+I)^(1/2)*(-I*x^2+x^2+1)^(1/2)*(I*x^2+x^2+1)^(1/2)/(2*x^4+2*x^2+1)^(1/2)*EllipticPi(x*(-1+I)^(1/2),1/3+1/3*I
,(-1-I)^(1/2)/(-1+I)^(1/2))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{4}}{\sqrt{2 \, x^{4} + 2 \, x^{2} + 1}{\left (2 \, x^{2} + 3\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(2*x^2+3)/(2*x^4+2*x^2+1)^(1/2),x, algorithm="maxima")

[Out]

integrate(x^4/(sqrt(2*x^4 + 2*x^2 + 1)*(2*x^2 + 3)), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{2 \, x^{4} + 2 \, x^{2} + 1} x^{4}}{4 \, x^{6} + 10 \, x^{4} + 8 \, x^{2} + 3}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(2*x^2+3)/(2*x^4+2*x^2+1)^(1/2),x, algorithm="fricas")

[Out]

integral(sqrt(2*x^4 + 2*x^2 + 1)*x^4/(4*x^6 + 10*x^4 + 8*x^2 + 3), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{4}}{\left (2 x^{2} + 3\right ) \sqrt{2 x^{4} + 2 x^{2} + 1}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4/(2*x**2+3)/(2*x**4+2*x**2+1)**(1/2),x)

[Out]

Integral(x**4/((2*x**2 + 3)*sqrt(2*x**4 + 2*x**2 + 1)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{4}}{\sqrt{2 \, x^{4} + 2 \, x^{2} + 1}{\left (2 \, x^{2} + 3\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(2*x^2+3)/(2*x^4+2*x^2+1)^(1/2),x, algorithm="giac")

[Out]

integrate(x^4/(sqrt(2*x^4 + 2*x^2 + 1)*(2*x^2 + 3)), x)